Can someone explain why $\log e^2$ is equal to $2$. 
This may sound like a basic question but why is log (e^2) = 2. I'm not very good with log questions.
 A: Here "$\log$" means $\log_e$.
In general, $\log_a b$ is the exponent $x$ such that $a^x=b$.
So, $\log_e e^2$ is the exponent $x$ such that $e^x=e^2$.
Evidently that is $2$.
A: If log stands for logarithm to base $e$ then $\log x=y$ is equivalent to $x=e^{y}$. From this definition it is obvious that $log (e^{2})=2$. 
A: Compute
$$
A := \int_1^{e^2}\frac{dt}{t}.
$$
Note $A = B+C$, where
$$
B= \int_1^{e}\frac{dt}{t}\qquad C=\int_e^{e^2}\frac{dt}{t}
$$
Of course $B = 1$, that is the definition of $e$.
For $C$, substitute $t=es,\; dt = e\;ds$ to see
$$
C = \int_e^{e^2}\frac{dt}{t} = \int_1^e \frac{e\;ds}{e\;s} = B = 1.
$$
Therefore $A = B+C = 1+1 = 2$.
A: First of all.  By $\log$ the mean natural log which is often (some could even argue usually) written as $\ln$.  It does not mean $\log_{10}$ or log base $10$ which is often (some could even argue usually) what $\log$ is assumed to mean.
Now $\log_b a$ (what $b > 0$ and $b \ne 1$ and $a > 0$ is defined to be "whatever it takes to get $b^{??????} = a$, then that value is what $\log_b a$ is defined to be.
So as a result $b^{\log_b a} =a$.  Always.
Basically just remember (assuming $b > 0$ and $b \ne 1$ and $a >0$) that $b^k =a \iff \log_b a = k$.  That's a definition.
So if $\log 2 = \log_e 2 = \log 2$ then $e^{\log 2} =2$ by definition.
.......
Sort of.
In calculus we usually start with defining $Ln(x) = \int_1^x \frac 1t dt$ and $exp(x)$ as the inverse of $Ln(x)$.  That is if $Ln(k) = x$ then $exp(x)$ is that value $k$ that makes $\int_1^k\frac 1t dt = x$.
Then $e = exp(1) = \lim(1+\frac 1n)^n$. and as it all works out (... ignore my crazy wandwaving...) that $e^x = exp(x)$ and $\log_e x = Ln(x) = \int_1^x \frac 1t dt$. and that way we can defint what $b^x$ means for irrational real $x$.
